What Is the Net Thermal Radiation Transfer Rate for a Stretched Cylinder?

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SUMMARY

The net thermal radiation transfer rate for a solid cylinder is calculated using the formula P = σεA(Tenv^4 - T^4). For a cylinder with a radius of 2.5 cm, length of 5.1 cm, emissivity of 0.90, and an environment temperature of 46°C, the net thermal radiation transfer rate P1 is determined to be 1.304 W. When the cylinder is stretched to a radius of 0.51 cm, the area calculation for the new configuration must consider the volume conservation, leading to a new height that affects the area used in the calculation of P2. The ratio P2/P1 is dependent on correctly determining the new dimensions of the cylinder after stretching.

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  • Knowledge of geometric formulas for solid cylinders
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VitaX
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Homework Statement



A solid cylinder of radius r1 = 2.5 cm, length h1 = 5.1 cm, emissivity 0.90, and temperature 28°C is suspended in an environment of temperature 46°C. (a) What is the cylinder's net thermal radiation transfer rate P1? (b) If the cylinder is stretched until its radius is r2 = 0.51 cm, its net thermal radiation transfer rate becomes P2. What is the ratio P2/P1?

Homework Equations



P = σεA(Tenv^4 - T^4)

The Attempt at a Solution



My answer to part A is correct. I got P = (5.67E-8)(.9)(.0119)(319^4 - 301^4) = 1.304 W

But my answer to part B is wrong apparently. I don't know what I did wrong. But the formula for Area of Solid Cylinder = 2*pi*r^2 + 2*pi*r*h

I used r1 = .025 m and h1 = .051 m for Area in part a. But in part B the only thing I changed was r2 = .0051 m to find the new area. Then got P2 and divided it by P1 to get the ratio. But it says part B is wrong. What exactly did I do wrong with the area in part b? Am I to assume h changes as well? My exact formula for part B Area is this: A = 2*pi*(.0051)^2 + 2*pi*(.0051)(.051) = .0018 m^2
 
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VitaX said:
But my answer to part B is wrong apparently. I don't know what I did wrong. But the formula for Area of Solid Cylinder = 2*pi*r^2 + 2*pi*r*h

I used r1 = .025 m and h1 = .051 m for Area in part a. But in part B the only thing I changed was r2 = .0051 m to find the new area.

If the cylinder was stretched, presumably it stretched in some direction...
Also presumably, the overall mass of the cylinder didn't change (it's still got the same amount of "stuff") so what does that say about the volume of the stretched cylinder?
 
gneill said:
If the cylinder was stretched, presumably it stretched in some direction...
Also presumably, the overall mass of the cylinder didn't change (it's still got the same amount of "stuff") so what does that say about the volume of the stretched cylinder?

So you are saying use V = pi*r^2*h to find the volume in part A. Then use that same volume and radius 2 in part B to find height 2. Then find the new area of part B using the new height and radius?
 
VitaX said:
So you are saying use V = pi*r^2*h to find the volume in part A. Then use that same volume and radius 2 in part B to find height 2. Then find the new area of part B using the new height and radius?

Sounds like a plan!
 

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